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Daniel [21]
3 years ago
7

A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current I. The current in the loop produces a magnetic field at th

e center of the loop that has magnitude 3.10×10−5 T and direction away from you as you view the plane of the loop. What are the magnitude and direction (clockwise or counterclockwise) of the current in the loop?
Physics
1 answer:
rusak2 [61]3 years ago
4 0

Answer:

1.63 A and in clockwise direction

Explanation:

The magnetic field due to the rectangular loop is :

$B=\frac{2 \mu_0 I}{\pi}\left(\frac{\sqrt{L^2+W^2}}{LW}\right)$

Given : W = 4.20 cm

                $=4.20 \times 10^{-2} \ m$

            L = 9.50 cm

               $= 9.50 \times 10^{-2} \ m$

            $B = 3.40 \times 10^{-5} \ T $

   Rearranging the above equation, we get

$I=\frac{B \pi LW}{2 \mu_0\sqrt{L^2+W^2}}$

$I=\frac{(3.40 \times 10^{-5}) \pi(9.50 \times 10^{-2})(4.20 \times 10^{-2})}{2(4 \pi \times 10^{-7})\sqrt{(9.50 \times 10^{-2})^2+(4.20 \times 10^{-2})^2}}$

I = 1.63 A

So the magnitude of the current in the rectangular loop is 1.63 A.

And the direction of current is clockwise.

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Answer:

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Explanation:

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\Delta U = Q-W

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In this problem,

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Therefore the work done by the system is

W=Q-\Delta U=757 kJ-176 kJ=+581 kJ

And the positive sign means the work is done BY the system.

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3 years ago
You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t
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Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together.  This is done because the friction force is going to have to be compensated for.  We will need that much more force than we otherwise would to achieve the desired acceleration:

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The friction force will be given by the normal force times the coefficient of friction.  Here the normal force is just its weight, mg

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Please help me i have this due tommorrow!!!
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8 0
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You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the
podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

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F_x = Fcos30


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N = Fsin30 + mg

N = 0.5F + (18\times 9.8)

N = 0.5F + 176.4

now the friction force on the cart is given as

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F_f = 0.625(0.5F + 176.4)

F_f = 110.25 + 0.3125F

now if cart moves with constant speed then net force on cart must be zero

so now we have

F_f + F_x = 0

Fcos30 - (110.25 + 0.3125F) = 0

0.866F - 0.3125F = 110.25

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In a series circuit what effect will adding more resistors to the circuit have
zhuklara [117]

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Explanation:

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